[proofplan]
The proof has two inputs. First, the local analytic Lojasiewicz inequality turns the hypothesis $f|_{Z(g)}=0$ into a local estimate $|f|^k \leq C|g|$. Second, the [effective Briançon-Skoda theorem](/theorems/3725) says that any [holomorphic function](/page/Holomorphic%20Function) satisfying this integral-closure estimate has its $(q+1)$st power in the ideal generated by $g_1,\dots,g_p$, where $q=\min(m,p-1)$. Applying this to $f^k$ on a smaller pseudoconvex neighbourhood gives the explicit exponent $k(q+1)$ and hence the local Nullstellensatz conclusion.
[/proofplan]
[step:Turn vanishing on the common zero set into a local Lojasiewicz estimate]
Fix $a \in \Omega$. We use the following local analytic Lojasiewicz inequality: if holomorphic functions $g_1,\dots,g_p,f$ are defined near $a$ and $f$ vanishes on the common zero set of $g_1,\dots,g_p$, then there exist an open neighbourhood $V \subset \Omega$ of $a$, an integer $k \geq 1$, and a constant $C>0$ such that
\begin{align*}
|f(z)|^k \leq C\left(\sum_{j=1}^p |g_j(z)|^2\right)^{1/2}
\end{align*}
for every $z \in V$.
Thus, for the fixed point $a$, we obtain $V$, $k$, and $C$ with
\begin{align*}
|f(z)|^k \leq C|g(z)|
\end{align*}
on $V$.
This cites a result not yet in the wiki: local analytic Lojasiewicz inequality.
[guided]
We first need to convert the geometric hypothesis $f=0$ on $Z(g)$ into an analytic estimate. The common zero set is
\begin{align*}
Z(g)=\{z \in \Omega : g_1(z)=\cdots=g_p(z)=0\}.
\end{align*}
The hypothesis says that $f(z)=0$ for every $z \in Z(g)$. The local analytic Lojasiewicz inequality applies to precisely this situation: holomorphic functions near a point, with one of them vanishing wherever the tuple $g=(g_1,\dots,g_p)$ vanishes.
Applying that inequality at the fixed point $a \in \Omega$, we obtain an open neighbourhood $V \subset \Omega$ of $a$, an integer $k \geq 1$, and a constant $C>0$ such that
\begin{align*}
|f(z)|^k \leq C\left(\sum_{j=1}^p |g_j(z)|^2\right)^{1/2}
\end{align*}
for every $z \in V$. Since we defined
\begin{align*}
|g(z)|:=\left(\sum_{j=1}^p |g_j(z)|^2\right)^{1/2},
\end{align*}
this is exactly
\begin{align*}
|f(z)|^k \leq C|g(z)|
\end{align*}
on $V$.
This cites a result not yet in the wiki: local analytic Lojasiewicz inequality.
[/guided]
[/step]
[step:Choose a pseudoconvex neighbourhood where Skoda division applies]
Choose an open polydisc $U \subset V$ with center $a$ and compact closure contained in $V$. Since every polydisc in $\mathbb{C}^m$ is pseudoconvex, $U$ is a pseudoconvex neighbourhood of $a$. Define the [holomorphic function](/page/Holomorphic%20Function)
\begin{align*}
\varphi: U &\to \mathbb{C} \\
z &\mapsto f(z)^k.
\end{align*}
The Lojasiewicz estimate on $V$ restricts to $U$ and gives
\begin{align*}
|\varphi(z)| \leq C|g(z)|
\end{align*}
for every $z \in U$.
[guided]
The [effective Briançon-Skoda theorem](/theorems/3725) is a local analytic division theorem, and it is normally applied on pseudoconvex domains. We therefore shrink the neighbourhood from $V$ to a standard pseudoconvex one. Choose an open polydisc $U \subset V$ centered at $a$ with compact closure contained in $V$. This is possible because $V$ is open and contains $a$. A polydisc is pseudoconvex, so $U$ has the geometric hypothesis required for Skoda division.
Now define
\begin{align*}
\varphi: U &\to \mathbb{C} \\
z &\mapsto f(z)^k.
\end{align*}
Because $f$ is holomorphic on $\Omega$ and $U \subset \Omega$, the function $\varphi$ is holomorphic on $U$. The Lojasiewicz estimate already obtained on $V$ remains valid after restriction to $U$, so for every $z \in U$,
\begin{align*}
|\varphi(z)|=|f(z)|^k \leq C|g(z)|.
\end{align*}
This is the integral-closure type bound needed for the Briançon-Skoda step.
[/guided]
[/step]
[step:Apply the effective Briançon-Skoda theorem to $f^k$]
Let $q:=\min(m,p-1)$. We use the [effective Briançon-Skoda theorem](/theorems/3725) in the following form: if $U \subset \mathbb{C}^m$ is pseudoconvex, $g_1,\dots,g_p \in \mathcal{O}(U)$, and $\varphi \in \mathcal{O}(U)$ satisfies $|\varphi| \leq C|g|$ on $U$ for some constant $C>0$, then
\begin{align*}
\varphi^{q+1} \in (g_1,\dots,g_p)\mathcal{O}(U).
\end{align*}
The hypotheses are satisfied: $U$ is pseudoconvex, each $g_j|_U$ is holomorphic, $\varphi=f^k|_U$ is holomorphic, and $|\varphi|\leq C|g|$ on $U$. Hence there exist holomorphic functions
\begin{align*}
h_j: U \to \mathbb{C}, \qquad 1 \leq j \leq p,
\end{align*}
such that
\begin{align*}
\varphi^{q+1}=\sum_{j=1}^p g_j h_j
\end{align*}
on $U$.
This cites a result not yet in the wiki: [effective Briançon-Skoda theorem](/theorems/3725).
[guided]
We now apply the division theorem that supplies ideal membership. The [effective Briançon-Skoda theorem](/theorems/3725) says the following. For a pseudoconvex domain $U \subset \mathbb{C}^m$, holomorphic generators $g_1,\dots,g_p \in \mathcal{O}(U)$, and a [holomorphic function](/page/Holomorphic%20Function) $\varphi \in \mathcal{O}(U)$ satisfying an estimate
\begin{align*}
|\varphi(z)| \leq C|g(z)|
\end{align*}
on $U$, the power $\varphi^{q+1}$ lies in the ideal generated by $g_1,\dots,g_p$, where
\begin{align*}
q=\min(m,p-1).
\end{align*}
We verify the hypotheses one by one. The set $U$ is pseudoconvex because it is a polydisc. Each restricted function
\begin{align*}
g_j|_U: U \to \mathbb{C}
\end{align*}
is holomorphic because $g_j$ is holomorphic on $\Omega$ and $U \subset \Omega$. The function
\begin{align*}
\varphi: U \to \mathbb{C}, \qquad z \mapsto f(z)^k,
\end{align*}
is holomorphic because it is a power of the [holomorphic function](/page/Holomorphic%20Function) $f|_U$. Finally, the estimate
\begin{align*}
|\varphi(z)| \leq C|g(z)|
\end{align*}
holds for every $z \in U$ by the previous step.
Therefore the [effective Briançon-Skoda theorem](/theorems/3725) gives holomorphic functions
\begin{align*}
h_j: U \to \mathbb{C}, \qquad 1 \leq j \leq p,
\end{align*}
such that
\begin{align*}
\varphi^{q+1}=\sum_{j=1}^p g_j h_j
\end{align*}
on $U$.
This cites a result not yet in the wiki: [effective Briançon-Skoda theorem](/theorems/3725).
[/guided]
[/step]
[step:Translate the Skoda conclusion into the asserted exponent]
Since $\varphi=f^k$ on $U$, the identity from the previous step becomes
\begin{align*}
f^{k(q+1)}=\sum_{j=1}^p g_j h_j
\end{align*}
on $U$. Thus
\begin{align*}
f^{k(q+1)} \in (g_1,\dots,g_p)\mathcal{O}(U).
\end{align*}
Consequently, when a local Lojasiewicz estimate with exponent $k$ is known, one may take $N=k(q+1)$. In particular, the Lojasiewicz estimate obtained from the vanishing hypothesis supplies some integer $k \geq 1$, so taking $N=k(q+1)$ proves the local Nullstellensatz assertion at the arbitrary point $a \in \Omega$.
[/step]
